## Precalculus (10th Edition)

In order for a sequence to be geometric, the quotient of all consecutive terms must be constant. Hence, in the given sequence we have: $\dfrac{a_2}{a_1}=\dfrac{\frac{3}{4}}{\frac{2}{3}}=\dfrac{9}{8}$ $\dfrac{a_3}{a_2}=\dfrac{\frac{4}{5}}{\frac{3}{4}}=\dfrac{16}{15}$ Since the quotient sare not the same, then the sequence is not geometric. In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant. Hence, here we have: $a_2-a_1=\dfrac{3}{4}-\dfrac{2}{3}=\dfrac{1}{12}$ $a_3-a_2=\dfrac{4}{5}-\dfrac{3}{4}=\dfrac{1}{20}$ The the difference is not constant, then the sequence us not an arithmetic sequence. Thus, the sequence us neither arithmetic nor geometric.